Question:
A U-tube manometer uses mercury \((\rho = 13,600\ \mathrm{kg/m^3})\). If the level difference is \(10\ \mathrm{cm}\), then the pressure difference is approximately
A U-tube manometer uses mercury \((\rho = 13,600\ \mathrm{kg/m^3})\). If the level difference is \(10\ \mathrm{cm}\), then the pressure difference is approximately
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For a U-tube manometer,
\[
\boxed{
\Delta P=\rho gh
}
\]
where \(h\) is the difference in liquid levels.
Updated On: Jul 14, 2026
- \(1.33\ \mathrm{kPa}\)
- \(6.67\ \mathrm{kPa}\)
- \(13.34\ \mathrm{kPa}\)
- \(26.68\ \mathrm{kPa}\)
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The Correct Option is C
Solution and Explanation
Step 1: Use the hydrostatic pressure relation.
The pressure difference measured by a U-tube manometer is
\[
\Delta P=\rho gh.
\]
Given,
\[
\rho=13600\ \mathrm{kg/m^3},
\]
\[
g=9.81\ \mathrm{m/s^2},
\]
\[
h=10\ \mathrm{cm}=0.1\ \mathrm{m}.
\]
Step 2: Calculate the pressure difference. \[ \Delta P = 13600\times9.81\times0.1 = 13341.6\ \mathrm{Pa} \] \[ =13.34\ \mathrm{kPa}. \] Hence, \[ \boxed{13.34\ \mathrm{kPa}} \] Therefore, \[ \boxed{(C)} \] is the correct answer.
Step 2: Calculate the pressure difference. \[ \Delta P = 13600\times9.81\times0.1 = 13341.6\ \mathrm{Pa} \] \[ =13.34\ \mathrm{kPa}. \] Hence, \[ \boxed{13.34\ \mathrm{kPa}} \] Therefore, \[ \boxed{(C)} \] is the correct answer.
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